package leetcode

import com.sun.org.apache.xpath.internal.operations.Bool
import kotlinetc.println

//https://leetcode.com/problems/camelcase-matching/

/**
A query word matches a given pattern if we can insert lowercase letters to the pattern word so that it equals the query. (We may insert each character at any position, and may insert 0 characters.)

Given a list of queries, and a pattern, return an answer list of booleans, where answer[i] is true if and only if queries[i] matches the pattern.



Example 1:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
Output: [true,false,true,true,false]
Explanation:
"FooBar" can be generated like this "F" + "oo" + "B" + "ar".
"FootBall" can be generated like this "F" + "oot" + "B" + "all".
"FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer".
Example 2:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
Output: [true,false,true,false,false]
Explanation:
"FooBar" can be generated like this "Fo" + "o" + "Ba" + "r".
"FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll".
Example 3:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
Output: [false,true,false,false,false]
Explanation:
"FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est".


Note:

1 <= queries.length <= 100
1 <= queries[i].length <= 100
1 <= pattern.length <= 100
All strings consists only of lower and upper case English letters.
 */
fun main(args: Array<String>) {

    camelMatch(arrayOf("FooBar", "FooBarTest", "FootBall", "FrameBuffer", "ForceFeedBack"), "FoBaT")
            .toList().println()
}

fun camelMatch(queries: Array<String>, pattern: String): BooleanArray {

    val result = BooleanArray(queries.size) { false }

    for (i in 0 until queries.size) {
        val query = queries[i].toCharArray()

        var j = 0
        var r = false
        for (k in 0 until query.size) {
            //模式比匹配串长，不匹配
            if (pattern.length > query.size) break

            if (j < pattern.length && query[k] == pattern[j]) {

                j++
            } else if (query[k] in 'A'..'Z') {//如果是 大写字母还不相等，那么说明不匹配
                r = true
                result[i] = false
                break
            }

            //遇到小写字母，query  k 继续战术后移
        }

        if (r.not())
            result[i] = j == pattern.length//应该是移动到最后一位的后面，刚好越界

    }

    return result
}


